A184633 Floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.

1, 4, 12, 28, 55, 96, 152, 227, 324, 444, 591, 768, 976, 1219, 1500, 1820, 2183, 2592, 3048, 3555, 4116, 4732, 5407, 6144, 6944, 7811, 8748, 9756, 10839, 12000, 13240, 14563, 15972, 17468, 19055, 20736, 22512, 24387, 26364, 28444, 30631, 32928, 35336, 37859, 40500, 43260, 46143, 49152, 52288, 55555, 58956, 62492, 66167, 69984, 73944, 78051, 82308, 86716, 91279, 96000, 100880, 105923, 111132, 116508, 122055, 127776, 133672, 139747, 146004, 152444, 159071, 165888, 172896, 180099, 187500

Offset: 1

Author

Clark Kimberling, Jan 18 2011

Keywords

• nonn

Links

• Edward Jiang and Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1000 terms from Edward Jiang)
• Index entries for linear recurrences with constant coefficients, signature (3, -3, 2, -3, 3, -1).

Crossrefs

Cf. A184536.

Programs

Mathematica

p[n_]:=FractionalPart[(n^4+9)^(1/4)]; q[n_]:=Floor[1/p[n]];
  Table[q[n], {n, 1, 80}]
  FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1,4},LinearRecurrence[{3,-3,2,-3,3,-1},{12,28,55,96,152,227},73]] (* Ray Chandler, Aug 02 2015 *)

PARI

a(n)=my(t=(9+n^4)^(1/4)); 1\(t-t\1) \\ Charles R Greathouse IV, Sep 12 2014

Formula

a(n) = floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.

It appears that a(n) = 3a(n-1)-3a(n-2)+2a(n-3)-3a(n-4)+3a(n-5)-a(n-6) for n>=9.

Empirical g.f.: x*(x+1)*(x^6-3*x^5+3*x^4-x^3+3*x^2+1) / ((x-1)^4*(x^2+x+1)). - Colin Barker, Jun 13 2015